Question
If $a, b, c$ are in continued proportion, prove that: $a : c = (a^2 + b^2) : (b^2 + c^2)$
✓
Answer
As $a, b, c$, are in continued proportion
$
\begin{aligned}
& \text { Let } \frac{a}{b}=\frac{b}{c}= k \\
& \text { a }: c =\left( a ^2+ b ^2\right):\left( b ^2+ c ^2\right) \\
& \Rightarrow \frac{a}{c}=\frac{a^2+b^2}{b^2+c^2} \\
& \text { L.H.S. }=\frac{a}{c} \\
& =\frac{c k^2}{c} \\
& = k ^2 \\
& \text { R.H.S. }=\frac{\left(c k^2\right)^2+(c k)^2}{(c k)^2+c^2} \\
& =\frac{c^2 k^4+c^2 k^2}{c^2 k^2+c^2} \\
& =\frac{c^2 k^2\left(k^2+1\right)}{c^2\left(k^2+1\right)} \\
& = k ^2 \\
& \therefore \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
$
\begin{aligned}
& \text { Let } \frac{a}{b}=\frac{b}{c}= k \\
& \text { a }: c =\left( a ^2+ b ^2\right):\left( b ^2+ c ^2\right) \\
& \Rightarrow \frac{a}{c}=\frac{a^2+b^2}{b^2+c^2} \\
& \text { L.H.S. }=\frac{a}{c} \\
& =\frac{c k^2}{c} \\
& = k ^2 \\
& \text { R.H.S. }=\frac{\left(c k^2\right)^2+(c k)^2}{(c k)^2+c^2} \\
& =\frac{c^2 k^4+c^2 k^2}{c^2 k^2+c^2} \\
& =\frac{c^2 k^2\left(k^2+1\right)}{c^2\left(k^2+1\right)} \\
& = k ^2 \\
& \therefore \text { L.H.S. }=\text { R.H.S. }
\end{aligned}
$
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